$w_1=48[\cos(\dfrac{19\pi}{20})+i\sin(\dfrac{19\pi}{20})]$ $w_2=16[\cos(\dfrac{\pi}{10})+i\sin(\dfrac{\pi}{10})]$ Express the quotient, $\dfrac{w_1}{w_2}$, in polar form. The angle should be given in radians. $\dfrac{w_1}{w_2}= $
Solution: Background For any two complex numbers $z_1$ and $z_2$ (whose radii are $r_1$ and $r_2$ and angles are $\theta_1$ and $\theta_2$ ): The radius of $\dfrac{z_1}{z_2}$ is the quotient of the original radii, $\dfrac{r_1}{r_2}$. The angle of $\dfrac{z_1}{z_2}$ is the difference of the original angles, $\theta_1 - \theta_2$. In other words, suppose the polar forms of $z_1$ and $z_2$ are as follows, $z_1 = r_1[\cos(\theta_1) + {i}\sin(\theta_1)]$ $z_2 = r_2[\cos(\theta_2) + {i}\sin(\theta_2)]$, then the polar form of their quotient is: $\dfrac{z_1}{z_2} = \dfrac{r_1}{r_2}[\cos(\theta_1 - \theta_2) + {i}\sin(\theta_1 - \theta_2)]$. [How do we get this?] Finding the radius of $\dfrac{w_1}{w_2}$ $w_1=48[\cos(\dfrac{19\pi}{20})+i\sin(\dfrac{19\pi}{20})]$ $w_2=16[\cos(\dfrac{\pi}{10})+i\sin(\dfrac{\pi}{10})]$ Here, $r_1=48$ and $r_2=16$. Therefore, the radius of $\dfrac{w_1}{w_2}$ is $\dfrac{r_1}{r_2}=3$. Finding the angle of $\dfrac{w_1}{w_2}$ $w_1=48[\cos(\dfrac{19\pi}{20})+i\sin(\dfrac{19\pi}{20})]$ $w_2=16[\cos(\dfrac{\pi}{10})+i\sin(\dfrac{\pi}{10})]$ Here, $\theta_1=\dfrac{19\pi}{20}$ and $\theta_2=\dfrac{\pi}{10}$. Therefore, the angle of $\dfrac{w_1}{w_2}$ is $\theta_1-\theta_2=\dfrac{19\pi}{20}-\dfrac{\pi}{10}=\dfrac{17\pi}{20}$ Summary We found that the radius of $\dfrac{w_1}{w_2}$ is $3$ and its angle is $\dfrac{17\pi}{20}$. Therefore, $\dfrac{w_1}{w_2}=3\left(\cos\left(\dfrac{17\pi}{20}\right)+i\sin\left(\dfrac{17\pi}{20}\right)\right)$